91影视

The given cylinder is$x^2+y^2=4$and the plane is$x-3y+2z=6.$

It is given that we have to find the point closest to the origin so, the distance of a point from the origin is $\sqrt{x^2+y^2+z^2}.$

Let the function be the square of the above distance $D(x,y,z)=x^2+y^2+z^2$and the constraint as $g(x,y,z)=x^2+y^2鈭�4=0\text{and}h(x,y,z)=x鈭�3y+2z鈭�6=0.$

By using the Lagrange's method, the gradient of the above functions is,

$\begin{array}{rl}\mathrm{鈭�}D(x,y,z)=& 2x\mathbf{i}+2y\mathbf{j}+2z\mathbf{k}\\ \mathrm{鈭�}g(x,y,z)=& 2x\mathbf{i}+2y\mathbf{j}\\ \mathrm{鈭�}h(x,y,z)=& \mathbf{i}鈭�3\mathbf{j}+2\mathbf{k}\end{array}$

Now, the system of the equation is

role="math" localid="1649984979254" $$\begin{gathered} \mathrm{鈭�}D(x,y,z)=位\mathrm{鈭�}g(x,y,z)+渭\mathrm{鈭�}h(x,y,z) \\ 2x\mathbf{i}+2y\mathbf{j}+2z\mathbf{k}=位(2x\mathbf{i}+2y\mathbf{j})+渭(\mathbf{i}鈭�3\mathbf{j}+2\mathbf{k}) \\ 2x\mathbf{i}+2y\mathbf{j}+2z\mathbf{k}=2位x\mathbf{i}+2位y\mathbf{j}+渭\mathbf{i}鈭�3渭\mathbf{j}+2渭\mathbf{k} \end{gathered}$$

Here,

$\begin{array}{rl}2x=& 位2x+渭\\ 2y=& 位2y鈭�3渭\\ 2z=& 0+2渭\end{array}$

It is clear that $z=渭.$

The values of $位,$we get are

$$\begin{gathered} 2x=位2x+z \\ 位=\frac{2x鈭�z}{2x}........\left(a\right) \\ 2y=位2y-3z \\ 位=\frac{2y+3z}{2y}...........\left(b\right) \end{gathered}$$

So, equate equations (a) and (b)

$$\begin{gathered} \frac{2x-z}{2x}=\frac{2y+3z}{2y} \\ \frac{2x-z}{x}=\frac{2y+3z}{y} \\ 2xy-yz=2xy+3xz \\ 3xz+yz=0 \\ z(3x+y)=0 \\ z=0y=-3x \end{gathered}$$

Substitute the above values in the given constraints,

$$\begin{gathered} \mathrm{If}z=0 \\ {x}^2+y^2=4....\left(1\right) \\ x-3y=6 \\ x=6+3y \\ \mathrm{Substitute}\mathrm{above}\mathrm{equation}\mathrm{in}\mathrm{equation}\left(1\right) \\ {\left(6+3y\right)}^2+{\mathrm{y}}^2=4 \\ 36+9{\mathrm{y}}^2+54\mathrm{y}=4 \\ 9{\mathrm{y}}^2+54\mathrm{y}+32=0 \\ \mathrm{y}=\frac{-54卤\sqrt{1764}}{18} \\ \mathrm{y}=\frac{-54卤42}{18} \\ \mathrm{y}=-\frac{2}{3},-\frac{43}{9} \end{gathered}$$

So, role="math" localid="1649986188071" $\mathrm{if}y=-\frac{2}{3},x=4\mathrm{and}\mathrm{if}y=-\frac{43}{9},x=-\frac{25}{3}.$

Thus, the points are $\left(4,-\frac{2}{3},0\right)\mathrm{and}\left(-\frac{25}{3},-\frac{43}{9},0\right).$

Now, let's substitute $y=-3x$in the given constraints,

$$\begin{gathered} x^2+{\left(-3x\right)}^2=4 \\ 4x^2=4 \\ x=卤1 \\ \mathrm{And} \\ x-3\left(-3x\right)+2z=6 \\ x+9x+2z=6 \\ 10x+2z=6 \\ 5x+z=3 \end{gathered}$$

The equation $5x+z=3$gives complex roots.

So, the points in${\mathrm{鈩�}}^3$closest to the origin are$\left(4,-\frac{2}{3},0\right)\mathrm{and}\left(-\frac{25}{3},-\frac{43}{9},0\right).$